Question

If 3 sin B + 5 cos B = 5, show that 3 cos B - 5 sin B = 3.

A. 3 sin B + 5 cos B = 5
B. 3 cos B - 5 sin B = 3
C. sin A
D. sin Atmos

Answer 1

To show that 3 cos B - 5 sin B = 3, we can start with the given equation 3 sin B + 5 cos B = 5. Multiply each term by 3 and rearrange the terms to get sin B = (1 - cos B) / 3. Use the identity sin(a + b) = sin a cos b + cos a sin b to rewrite the equation and simplify to show that 3 cos B - 5 sin B = 3.

Step-by-step explanation:

To show that 3 cos B - 5 sin B = 3, we can start with the given equation 3 sin B + 5 cos B = 5.

Let's multiply each term in the equation by 3: 3(3 sin B + 5 cos B) = 3(5). This gives us 9 sin B + 15 cos B = 15.

Now, rearrange the terms: 9 sin B = 15 - 15 cos B. Divide both sides by 9: sin B = (15 - 15 cos B) / 9.

Next, divide both the numerator and denominator on the right side of the equation by 15: sin B = (1 - cos B) / 3.

Finally, using the identity sin(a + b) = sin a cos b + cos a sin b, we can rewrite the equation as sin B = sin B cos B / 3 + cos B sin B / 3.

Simplifying further, we get sin B = (sin B cos B + cos B sin B) / 3.

This equation is true, so we have shown that 3 sin B + 5 cos B = 5 implies 3 cos B - 5 sin B = 3.